Problem 48 - Precalculus with Calculus Previews

Question

Find an equation of a sphere that satisfies the given conditions. $$ \text { Center }(0,-3,0) ; \text { diameter } \frac{5}{2} $$

Step-by-Step Solution

Answer

The equation of the sphere is \( x^2 + (y+3)^2 + z^2 = \frac{25}{16} \).

Step 1: Determine the Radius

The diameter of the sphere is given as \( \frac{5}{2} \). The radius \( r \) is half of the diameter, so:\[ r = \frac{5}{2} \div 2 = \frac{5}{4} \]

Step 2: Use the Sphere Equation Formula

The formula for the equation of a sphere with center \((h, k, l)\) and radius \(r\) is:\[ (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \]

Step 3: Substitute the Center Coordinates

Given the center of the sphere is \( (0, -3, 0) \), substitute \( h = 0 \), \( k = -3 \), and \( l = 0 \) into the equation:\[ (x-0)^2 + (y+3)^2 + (z-0)^2 = r^2 \]

Step 4: Substitute the Radius

Substitute the radius \( r = \frac{5}{4} \) into the equation and simplify:\[ x^2 + (y+3)^2 + z^2 = \left( \frac{5}{4} \right)^2 \]

Step 5: Square the Radius

Calculate \( \left( \frac{5}{4} \right)^2 \):\[ \left( \frac{5}{4} \right)^2 = \frac{25}{16} \]

Step 6: Write the Final Equation

Combine everything into the final equation of the sphere:\[ x^2 + (y+3)^2 + z^2 = \frac{25}{16} \]

Key Concepts

Radius of a Sphere

The radius of a sphere is a crucial concept for understanding spheres, as it directly influences the sphere's size. The radius (\( r \)) is the distance from the exact center of the sphere to any point on its surface.
In our case, to find the radius for the example sphere, we begin with the diameter, which is \( \frac{5}{2} \). Since the radius is half of the diameter, we calculate:

• \( r = \frac{5}{2} \div 2 = \frac{5}{4} \)

Understanding this helps in visualizing the sphere and is essential for solving problems involving sphere equations.
By knowing the radius, you can also calculate other characteristics of the sphere, such as its surface area or volume, using the respective formulas.

Center of a Sphere

The center of a sphere is another key element and is denoted by a point with coordinates \( (h, k, l) \). It is the "middle" point from which the sphere extends outward in all directions equally.
For the given problem, the center of the sphere is at \( (0, -3, 0) \). This means:

• The center lies directly on the y-axis, 3 units below the xz-plane due to the negative y-coordinate.
• The x and z coordinates being zero indicate it's aligned along the yz-plane.

Knowing the center is vital when using the equation of the sphere, as it shifts the origin of your coordinate system to the center of the sphere, helping to accurately describe its geometry in space.

Diameter of a Sphere

The diameter of a sphere is an important dimension, double the radius, stretching from one point on the sphere's surface, passing through the center, to the opposite point.
In the provided exercise, the diameter is given as \( \frac{5}{2} \). From this, we can deduce:

• That the radius is \( \frac{5}{4} \), as diameters are always twice the radius.
• Diameters help in visualizing the full extent of a sphere from the center point to its outer edge.

Having the diameter enables us to find other sphere parameters quickly and is essential in sphere-related calculations, like finding the volume or setting up its equation.